A solid ball of radius R has a charge density given by = 0 ( 1 - r/R) for 0 r R. The electric field outside the ball is

(A) \frac{{{\rho _0}{R^3}}}{{{\varepsilon _0}{r^2}}}

(B) \frac{{{4\rho _0}{R^3}}}{{{3\varepsilon _0}{r^2}}}

(C) \frac{{{3\rho _0}{R^3}}}{{{4\varepsilon _0}{r^2}}}

JEE Mains · Physics · STD 12 - Electric charges and fields

A solid ball of radius \(R\) has a charge density \(\rho \) given by \(\rho = {\rho _0}\left( {1 - \frac{r}{R}} \right)\) for \(0 \leq r \leq R\). The electric field outside the ball is

Pick an option to check your answer.

Verified Solution

Correct Answer

(D) \(\frac{{{\rho _0}{R^3}}}{{{12\varepsilon _0}{r^2}}}\)

Detailed explanation

\(\text { Charge density, } \rho=\rho_{0}\left(1-\frac{r}{R}\right)\)

\(d q=\rho d v\)

\(q_{i n}=\int d q=\rho d v\)

\(=\rho_{0}\left(1-\frac{r}{R}\right) 4 \pi r^{2} d r \quad\left(\because d v=4 \pi \mathrm{r}^{2} \mathrm{dr}\right)\)

\(=4 \pi p_{0} \int_{0}^{R}\left(1-\frac{r}{R}\right) r^{2} d r\)

\(=4 \pi \rho_{0} \int_{0}^{R} r^{2} d r-\frac{r^{2}}{R} d r\)

\(=4 \pi \rho_{0}\left[\left[\frac{r^{3}}{3}\right]_{0}^{R}-\left[\frac{r^{4}}{4 R}\right]_{0}^{R}\right]\)

\({=4 \pi \rho_{0}\left[\frac{R^{3}}{3}-\frac{R^{4}}{4 R}\right]}\)

\({=4 \pi \rho_{0}\left[\frac{R^{3}}{3}-\frac{R^{3}}{4}\right]=4 \pi \rho_{0}\left[\frac{R^{3}}{12}\right]}\)

\({q=\frac{\pi \rho_{0} R^{3}}{3}}\)

\(E .4 \pi r^{2}=\left(\frac{\pi \rho_{0} R^{3}}{3 \epsilon_{0}}\right)\)

Electric field outside the ball, \(E=\frac{\rho_{0} R^{3}}{12 \epsilon_{0} r^{2}}\)

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